Problem: Simplify and expand the following expression: $ \dfrac{1}{2t + 8}+ \dfrac{3}{4t + 32}- \dfrac{2}{t^2 + 12t + 32} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2t + 8} = \dfrac{1}{2(t + 4)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4t + 32} = \dfrac{3}{4(t + 8)}$ We can factor the quadratic in the third term: $ \dfrac{2}{t^2 + 12t + 32} = \dfrac{2}{(t + 4)(t + 8)}$ Now we have: $ \dfrac{1}{2(t + 4)}+ \dfrac{3}{4(t + 8)}- \dfrac{2}{(t + 4)(t + 8)} $ The least common multiple of the denominators is: $ 8(t + 4)(t + 8)$ In order to get the first term over $8(t + 4)(t + 8)$ , multiply by $\dfrac{4(t + 8)}{4(t + 8)}$ $ \dfrac{1}{2(t + 4)} \times \dfrac{4(t + 8)}{4(t + 8)} = \dfrac{4(t + 8)}{8(t + 4)(t + 8)} $ In order to get the second term over $8(t + 4)(t + 8)$ , multiply by $\dfrac{2(t + 4)}{2(t + 4)}$ $ \dfrac{3}{4(t + 8)} \times \dfrac{2(t + 4)}{2(t + 4)} = \dfrac{6(t + 4)}{8(t + 4)(t + 8)} $ In order to get the third term over $8(t + 4)(t + 8)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{2}{(t + 4)(t + 8)} \times \dfrac{8}{8} = \dfrac{16}{8(t + 4)(t + 8)} $ Now we have: $ \dfrac{4(t + 8)}{8(t + 4)(t + 8)} + \dfrac{6(t + 4)}{8(t + 4)(t + 8)} - \dfrac{16}{8(t + 4)(t + 8)} $ $ = \dfrac{ 4(t + 8) + 6(t + 4) - 16} {8(t + 4)(t + 8)} $ Expand: $ = \dfrac{4t + 32 + 6t + 24 - 16}{8t^2 + 96t + 256} $ $ = \dfrac{10t + 40}{8t^2 + 96t + 256}$ Simplify: $ = \dfrac{5t + 20}{4t^2 + 48t + 128}$